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Question: An electron (mass = \(9.1 \times 10^{- 31}kg\) and charge =\(1.6 \times 10^{- 19}coul.)\) is sent i...

An electron (mass = 9.1×10−31kg9.1 \times 10^{- 31}kg and charge

=1.6×10−19coul.)1.6 \times 10^{- 19}coul.) is sent in an electric field of intensity 1×106V/m.1 \times 10^{6}V/m. How long would it take for the electron, starting from rest, to attain one–tenth the velocity of light

A

1.7×10−12sec⁡1.7 \times 10^{- 12}\sec

B

1.7×10−6sec⁡1.7 \times 10^{- 6}\sec

C

1.7×10−8sec⁡1.7 \times 10^{- 8}\sec

D

1.7×10−10sec⁡1.7 \times 10^{- 10}\sec

Answer

1.7×10−6sec⁡1.7 \times 10^{- 6}\sec

Explanation

Solution

By using v=QEtmv = \frac{QEt}{m}

⇒ 110×3×108=(1.6×10−19)×106×t9.1×10−31\frac{1}{10} \times 3 \times 10^{8} = \frac{(1.6 \times 10^{- 19}) \times 10^{6} \times t}{9.1 \times 10^{- 31}}

⇒ t=1.7×10−10sec.t = 1.7 \times 10^{- 10}sec.